Chapter 12.2, Problem 1CE

(a)

To determine

To find: The volume of cube.

Answer to Problem 1CE

The volume of each cube is equal to $V=\frac{1}{6}{e}^3$ .

Explanation of Solution

Given information:

The diagonal of a cube intersect to divide the cube into six congruent pyramids as shown. The base of each pyramid is a face of the cube, and height of each pyramid is $\frac{1}{2}e$ .

  

Calculation:

The edge length of cube is equal to $e$ . The volume of cube is $e^3$ .

It is given that it is divided into six congruent pyramids. All of the six pyramids are similar to each other. So,

  $\begin{array}{c}6\left(\text{Volume}\text{of}\text{pyramid}\right)=e^3\\ \text{Volume}\text{of}\text{pyramid}=\frac{1}{6}{e}^3\end{array}$

Therefore, the volume of each cube is equal to $V=\frac{1}{6}{e}^3$ .

(b)

To determine

To prove: The volume of pyramid is $V=\frac{1}{3}Bh$ .

Explanation of Solution

Given information:

The diagonal of a cube intersect to divide the cube into six congruent pyramids as shown. The base of each pyramid is a face of the cube, and height of each pyramid is $\frac{1}{2}e$ .

  

Proof:

The base of each pyramid is $e$ and height of the pyramid is $\frac{1}{2}e$ .

From part (a), the volume of a pyramid is $V=\frac{1}{6}{e}^3$ .

Now, calculate the volume of pyramid in terms of base area as:

  $\begin{array}{l}V=\frac{1}{3}\left(e^2\right)\left(\frac{1}{2}e\right)\\ V=\frac{1}{3}Bh\end{array}$

Hence, it is proved that the volume of the pyramid is $V=\frac{1}{3}Bh$ .